First slide

Planes in 3D

Question

$The projection of the line \frac{x + 1}{- 1} = \frac{y}{2} = \frac{z - 1}{3} on the plane x - 2 y + z = 6 is the line of$ intersection of this plane with the plane

Moderate

A

$2 x + y + 2 = 0$
B

$3 x + y - z = 0$
C

$2 x - 3 y + 8 z = 3$
D

None of these

Solution

$Given line \frac{x + 1}{- 1} = \frac{y}{2} = \frac{z - 1}{3}$
$And plane is x - 2 y + z = 6$
$Equation of the plane through$
$(- 1, 0, 1) is a (x + 1) + b (y - 0) + c (z - 1) = 0 \dots \dots . (1)$
$Which is parallel to the given line and perpendicular to the given plane$
$\Rightarrow - a + 2 b + 3 c = 0$
$and a - 2 b + c = 0$
$\Rightarrow c = 0, a = 2 b$
$∴ From (1) equation of the required plane is 2 x + y + 2 = 0$

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